原题链接在这里:https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/description/ 题目: You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a sepa…

[LeetCode]430. Flatten a Multilevel Doubly Linked List 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/description/ 题目描述: You are given…

题目链接:LeetCode 430. Faltten a Multilevel Doubly Linked List class Node { public: int val = NULL; Node* prev = NULL; Node* next = NULL; Node* child = NULL; Node() {} Node(int _val, Node* _prev, Node* _next, Node* _child) { val = _val; prev = _prev; nex…

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so o…

/* // Definition for a Node. class Node { public: int val = NULL; Node* prev = NULL; Node* next = NULL; Node* child = NULL; Node() {} Node(int _val, Node* _prev, Node* _next, Node* _child) { val = _val; prev = _prev; next = _next; child = _child; } }…

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so o…

您将获得一个双向链表,除了下一个和前一个指针之外,它还有一个子指针,可能指向单独的双向链表.这些子列表可能有一个或多个自己的子项,依此类推,生成多级数据结构,如下面的示例所示. 扁平化列表,使所有结点出现在单级双链表中.您将获得列表第一级的头部. You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which m…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Hints: If you notice carefully in the flattened tree, each node's right…

题目 给定一个二叉树,原地将它展开为链表. 例如,给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4 \ 5 \ 6 解析 通过递归实现:可以用先序遍历,然后串成链表 主要思想就是:先递归对右子树进行链表化并记录,然后将root->right指向 左子树进行链表化后的头结点,然后一直向右遍历子树,连接上之前的右子树 /** * Definition for a binary tree node. * struct TreeNode { * int v…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 给一个二叉树,把它展平为链表 in-place 根据展平后的链表的顺序可以看出是先序遍历的结果,所以用inorder traversal. 解…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6   The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Hints: If you notice carefully in the flattened tree, each node's right child points to th…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 思路: 用先序遍历,得到的是从小到大的顺序.把先序遍历变形一下: void flatten(TreeNode* root) { if(NULL == root) return; vec…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 这道题就是将数变为只有右节点的树. 递归,不是很难. 递归的时候求出左节点的最右孩子即可. /** * Definition for a binary tree node. * pub…

随笔一记,留做重温! Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 第一个想法是先序遍历,然后按照访问顺序,添加右结点. public static void…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 题目 将左子树所形成的链表插入到root和root->right之间 思路 1 / 2(root) 假设当前root为2 / \ 3(p) 4…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 这个题思路就是DFS, 先左后右, 记住如果是用stack如果需要先得到左, 那么要先append右, 另外需要注意的就是用recursive…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Hints: If you notice carefully in the flattened tree, each node's right child points to the…

Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6解题思路:试图通过排序后new TreeNode是无法通过的,这道题的意思是把现有的树进行剪枝操作.JAVA实现如下: static public void flatten(TreeN…

Flatten Binary Tree to Linked List Total Accepted: 25034 Total Submissions: 88947My Submissions Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5…

Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 思路:这题主要是就是前序遍历.主要解法就是将左子树转换为右支树,同一时候加入在右子树前.左子树求解时,须要主要左子树的深度. 详细代码例如以下: /** * Definition f…

根据提示,本题等价于pre order traverse遍历,并且依次把所有的节点都存成right child,并把left child定义成空集.用递归的思想,那么如果分别把左右子树flatten成list,我们有: 1 /    \ 2     5 \       \ 3      6 <- rightTail \ 4  <- leftTail 所以使用递归的解法一: 注意由于右子树最后要接到左子树的后面,所以用temp保存右子树的head. def flatten(self, root)…

将左子树接到右子树之前,递归解决 void flatten(TreeNode *root) { if (root == nullptr)return; flatten(root->left); flatten(root->right); //如果没有左子树,直接返回即可 if (root->left == nullptr)return; p = root->left; //寻找左子树的最后一个结点 while (p->right)p = p->right; //将右结点…

把二叉树先序遍历,变成一个链表,链表的next指针用right代替 用递归的办法先序遍历,递归函数要返回子树变成链表之后的最后一个元素 class Solution { public: void helper(TreeNode* cur, TreeNode*& tail){ //a(tail) //lk("root",tail) //a(cur) //lk("root",cur) //dsp tail=cur; TreeNode* right=cur->…

/* 先序遍历构建链表,重新构建树 */ LinkedList<Integer> list = new LinkedList<>(); public void flatten(TreeNode root) { preOrder(root); TreeNode res = root; list.poll(); while (!list.isEmpty()) { res.right = new TreeNode(list.poll()); res.left = null; res =…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Hints: If you notice carefully in the flattened tree, each node's right…

题目来源 https://leetcode.com/problems/flatten-binary-tree-to-linked-list/ Given a binary tree, flatten it to a linked list in-place. 题意分析 Input: binary tree Output: flattened tree Conditions:将一个二叉树压平为一个flatten 树,也就是一条斜线 题目思路 先将左右子树压平,然后将左子树嵌入到本节点与右子树之间.…

Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Hints: If you notice carefully in the flattened tree, each node's right child points to the…

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Convert a BST to a sorted circular doubly-linked list in-place. Think of the left and right pointers as synonymous to the previous and next pointers in a doubly-linked list. Let's take the following BST as an example, it may help you understand the p…

题目说明 Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 题目分析 第一感觉是前序遍历,顺便打算在这题练习一下昨天学到的二级指针的写法XD,调的时候bug挺多的,可读性贼差,指针还是慎用啊-- 以下为个人实现(C++,12ms):…